De Moivre's Theorem states that for whole number n, $$(\cos \theta + i \sin \theta)^n = \cos n \theta + i \sin n \theta. $$ We can use this fact to derive certain trig identites: an example of a use of complex numbers to do real calculations that would otherwise be more difficult. Here's an example, in which we express $\cos 5 \theta$ in terms of powers of $\cos \theta. $
\begin{eqnarray} \cos 5 \theta + i \sin 5 \theta & = & (\cos \theta + i \sin \theta)^5 \\ & = & \cos^5 \theta + 5 i \cos^4 \theta \sin \theta + 10 i^2 \cos^3 \theta \sin^2 \theta + \\&&\quad10 i^3 \cos^2 \theta \sin^3 \theta + 5 i^4 \cos \theta \sin^4 \theta + i^5 \sin^5 \theta \\ & = & [\cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta] + \\&&\quad i [5 \cos^4 \theta \sin \theta - 10 \cos^2 \theta \sin^3 \theta + \sin^5 \theta]. \end{eqnarray}
Comparing real parts, $$\cos 5 \theta = \cos^5 \theta - 10 \cos^3 \theta \sin^2 \theta + 5 \cos \theta \sin^4 \theta.$$ Using the substitution \sin^2 \theta = 1 - \cos^2 \theta,
\begin{eqnarray} \cos 5 \theta & =& \cos^5 \theta - 10 \cos^3 \theta (1 - \cos^2 \theta) + 5 \cos \theta (1 - \cos^2 \theta)^2 \\ & = & 16 \cos^5 \theta - 20 \cos^3 \theta + 5 \cos \theta. \end{eqnarray}